Hydrodynamics for SSEP with Non-reversible Slow Boundary Dynamics: Part I, the Critical Regime and Beyond

Abstract

The purpose of this article is to provide a simple proof of the hydrodynamic and hydrostatic behavior of the SSEP in contact with slowed reservoirs which inject and remove particles in a finite size windows at the extremities of the bulk. More precisely, the reservoirs inject/remove particles at/from any point of a window of size K placed at each extremity of the bulk and particles are injected/removed to the first open/occupied position in that window. The hydrodynamic limit is given by the heat equation with non-linear Robin boundary conditions or Neumann boundary conditions, the latter being in the case when the reservoirs are too slow. The proof goes through the entropy method of Guo et al. (Commun Math Phys 118:31–59, 1988). We also derive the hydrostatic limit for this model, whose proof is based on the method developed in Landim and Tsunoda (Ann Henri Poincaré 54(1):51–74, 2018) and Tsunoda (Hydrostatic limit for exclusion process with slow boundary revisited, RIMS Kôkyûroku Bessatsu). We observe that we do not make use of correlation estimates in none of our results.

Appendices

Appendix A: Replacement Lemmas

In this section we prove the replacements lemmas that are needed along the arguments presented above. We start by obtaining an estimate relating the Dirichlet form and the carré du champ operator for this model. As above, for simplicity of the presentation, we state and prove the results for the case $K=2$, but the extension to the general case is completely analogous.

1.1 A.1. Dirichlet Forms

Let $\rho :[0,1]\rightarrow [0,1]$ be a measurable profile and let $\nu ^{N}_{\rho (\cdot )}$ be the Bernoulli product measure on $\Omega _{N}$ defined by

$$\begin{aligned} \nu ^{N}_{\rho (\cdot )}(\eta :\eta (x)=1)= \rho (\tfrac{x}{N}). \end{aligned}$$

(38)

For a probability measure $\mu$ on $\Omega _{N}$ and a density $f:\Omega _{N} \rightarrow \mathbb {R}$ with respect to $\mu$, the Dirichlet form is defined as

$$\begin{aligned} \langle f, -{\mathcal L}_N f\rangle _\mu = \langle f, -{\mathcal L}_{N,0}f \rangle _{\mu }+{\tfrac{1}{N^\theta }}\langle f, -{\mathcal L}_{N,b}f \rangle _{\mu }, \end{aligned}$$

(39)

and the carré du champ is defined by:

$$\begin{aligned} D_N(\sqrt{f}, \mu ):=D_{N,0}(\sqrt{f},\mu )+ {\tfrac{1}{N^\theta }}D_{N,b}(\sqrt{f},\mu ), \end{aligned}$$

(40)

where

$$\begin{aligned} D_{N,0}(\sqrt{f},\mu )\;:= & {} \;\sum _{x=1}^{n-2} \int _{\Omega _N}\left[ \sqrt{f(\eta ^{x,x+1})}- \sqrt{f(\eta )}\right] ^{2} d\mu ,\nonumber \\ D_{N,\pm }(\sqrt{f},\mu )= & {} \sum _{x\in I_\pm ^K}\int c_x^\pm (\eta ) \big [\sqrt{f(\eta ^{x})}-\sqrt{f(\eta )}\big ]^{2}\,d\mu , \end{aligned}$$

(41)

where we recall the rates $c^\pm _x$ defined in (5) , and $D_{N,b}=D_{N,-}+D_{N,+}$. We claim that for $\theta \geqslant 1$ and for $\rho :[0,1]\rightarrow [0,1]$ a constant profile equal to, for example, $\alpha$, the following bound holds

$$\begin{aligned} \begin{aligned} \langle {{\mathcal L}}_N\sqrt{f},\sqrt{f} \rangle _{\nu ^N_\alpha }&\lesssim -D_N(\sqrt{f},\nu ^N_\alpha ) + O(\tfrac{1}{ N}). \end{aligned} \end{aligned}$$

(42)

From Lemmas 5.1 and 5.2 of [1] it is only necessary to control the contribution from the non-linear part of the boundary dynamics. To do that, it is enough to apply Lemma 5.1 of [3] and the result follows. We leave these computations to the reader.

Remark A.1

(On the bound of the Dirichlet form) Note that for any a, b we have the identity $ab-b^2=-\tfrac{1}{2} (a-b)^2+\tfrac{1}{2} (a^2-b^2)$. This implies directly that

$$\begin{aligned} \langle \mathcal {L}_N\sqrt{f},\sqrt{f} \rangle _{\mu }=-\frac{1}{2}D_N(\sqrt{f},\mu )+\frac{1}{2}E_{\mu }\left[ (\mathcal {L}_Nf)(\eta )\right] \end{aligned}$$

for any measure $\mu$. If $\mu$ is the invariant measure of the system, then the last term on the right hand side of last display vanishes. For this model we have no information about the stationary measure and by taking $\mu =\nu _\alpha ^N$ this term is non zero and has to be controlled. For the bulk dynamics, when computing that term with respect to the $\nu _\alpha ^N$, it clearly vanishes, but the same is not true for the boundary dynamics. Since we are dealing with the case $\theta \geqslant 1$ we are reduced to bound $N^{-\theta }E_{\nu _\alpha ^N}[(\mathcal {L}_{N,b}f)(\eta )]$ and in this case the bounds provided by Lemma 5.1 of [3] are enough. Nevertheless, in the case $\theta <1$ these bounds do not work any more and a new argument is needed.

1.2 A.2. Replacement Lemmas

We start this section by proving the next lemma which is the basis for the replacement lemmas that are presented next.

Lemma A.2

Let $x<y\in \Lambda _N$ and let $\varphi :\Omega _N\rightarrow \Omega _N$ be a bounded function which satisfies $\varphi (\eta )=\varphi (\eta ^{z,z+1})$ for any $z=x,\cdots , y-1$. For any density f with respect to $\nu _{\alpha }$ and any positive constant A, it holds that

$$\begin{aligned} \left| \langle \varphi (\eta )( \eta (x)-\eta (y)), f \rangle _{\nu ^N_\alpha } \right| \lesssim \tfrac{1}{A} D_N(\sqrt{f},\nu ^N_\alpha ) + A(y-x). \end{aligned}$$

Proof

By summing and subtracting appropriate terms, we have that

$$\begin{aligned} \begin{aligned}&\vert \langle \varphi (\eta )( \eta (x)-\eta (y)),f \rangle _{\nu ^N_\alpha } \vert \\&\quad \leqslant \dfrac{1}{2} \sum _{z=x}^{y-1}\left| \int \varphi (\eta )( \eta (z)-\eta (z+1))[f(\eta )-f(\eta ^{z,z+1})] \; d\nu ^N_\alpha \right| \\&\qquad + \dfrac{1}{2}\sum _{z=x}^{y-1} \left| \int \varphi (\eta )( \eta (z) \right. \\&\qquad \left. -\eta (z+1))[f(\eta )+f(\eta ^{z,z+1})] \; d\nu ^N_\alpha \right| . \end{aligned} \end{aligned}$$

Note that since $\varphi$ satisfies $\varphi (\eta )=\varphi (\eta ^{z,z+1})$ for any $z=x,\cdots , y-1$, by a change of variables, we conclude that the last term in the previous display is equal to zero. Now, we treat the remaining term. Using the equality $(a-b)=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ and then Young’s inequality, the first term at the right-hand side of last display is bounded from above by a constant times

$$\begin{aligned} \begin{aligned}&\sum _{z=x}^{y-1}{\frac{A}{4}}\int (\varphi (\eta )(\eta (z)-\eta (z+1)))^2\left( \sqrt{f(\eta ^{z,z+1})}+ \sqrt{f(\eta )}\right) ^{2} d\nu ^N_\alpha +\dfrac{1}{{4A}}D_N(\sqrt{f},\nu ^N_\alpha ). \end{aligned} \end{aligned}$$

The fact that $\varphi$ is bounded, $|\eta (x)|\leqslant 1$ and f is a density, the integral in last expression is bounded from above by a constant. This ends the proof.$\square$

We are now able to show the first Replacement Lemma.

Lemma A.3

Fix $x,y\in \Lambda _N$ such that $|x-y|=o(N)$. Let $\varphi :\Omega _N\rightarrow \Omega _N$ be a bounded function which satisfies $\varphi (\eta )=\varphi (\eta ^{z,z+1})$ for any $z=x,\cdots , y-1$. For any $t\in [0,T]$ we have that

$$\begin{aligned} \begin{aligned}&\displaystyle \limsup _{N\rightarrow +\infty } \mathbb {E}_{\mu _{N}}\left[ \, \left| \displaystyle \int _{0}^{t}\varphi (\eta _{sN^2})(\eta _{sN^2}(x)- \eta _{sN^{2}}(y))ds \right| \,\right] = 0. \end{aligned} \end{aligned}$$

(43)

Proof

The starting point in the proof is to change from the measure $\mu _N$ to a suitable measure, which for our purposes is the Bernoulli product measure $\nu _{\rho (\cdot )}^N$ with a constant profile $\rho (\cdot )=\alpha \in (0,1)$. By the explicit formula for the entropy, it holds

$$\begin{aligned} \begin{aligned} H(\mu _N|\nu _{\alpha }^N)&=\sum _{\eta \in \Omega _N} \mu _N(\eta )\log \bigg (\frac{\mu _N(\eta )}{\nu _{\rho (\cdot )}^N(\eta )}\bigg ) \leqslant N \log \bigg (\frac{1}{C_{\alpha }}\bigg )\sum _{\eta \in \Omega _N} \mu _N(\eta )=C_{\alpha } N. \end{aligned} \end{aligned}$$

Therefore, by the entropy inequality and Jensen’s inequality, for any $B>0$ the expectation in the statement of the lemma is bounded by

$$\begin{aligned} \frac{C_\alpha }{B} + \frac{1}{NB} \log {\mathbb {E}}_{\nu _{\alpha }^N}\bigg [ e^{|\int _0^t BN\varphi (\eta _{sN^2})(\eta _{sN^2}(x)- \eta _{sN^{2}}(y))\,ds| } \bigg ]. \end{aligned}$$

(44)

Since $e^{|x|}\leqslant e^x+e^{-x}$ and

$$\begin{aligned} \underset{N\rightarrow \infty }{\limsup }\, \tfrac{1}{N}\log (a_N+b_N)\leqslant \max \big \{\underset{N\rightarrow \infty }{\limsup }\, \tfrac{1}{N}\log (a_N), \underset{N\rightarrow \infty }{\limsup }\, \tfrac{1}{N}\log (b_N) \big \}, \end{aligned}$$

we can remove the absolute value from (44). By Feynman-Kac’s formula (see Lemma 7.3 in [1]), (44) is bounded by

$$\begin{aligned} \frac{C_{\alpha }}{B} + t\sup _{f} \Big \{ |\langle \varphi (\eta )(\eta (x)-\eta (y)), f \rangle _{\nu ^N_\alpha } |+ \tfrac{N}{B} \langle {\mathcal L}_{N}\sqrt{f},\sqrt{f} \rangle _{\nu ^N_\alpha } \Big \}, \end{aligned}$$

where the supremum above is over densities f with respect to $\nu _\alpha ^N$. By Lemma A.2 with the choice $A=\tfrac{B}{N}$ we have that

$$\begin{aligned} \left| \langle \varphi (\eta )( \eta (x)-\eta (y)), f \rangle _{\nu ^N_\alpha } \right| \lesssim \tfrac{N}{B} D_{N}(\sqrt{f},\nu ^N_\alpha ) + \tfrac{B}{N}{|y-x|}{.} \end{aligned}$$

From (42) and the inequality above, the term on the right-hand side of (44), is bounded from above by $\tfrac{B}{N}{|y-x|} +{\tfrac{1}{B}}.$ Taking $N \rightarrow \infty$ and then $B\rightarrow +\infty$ we are done.$\square$

Lemma A.4

(Replacement Lemma) Let $\psi :\Omega _N\rightarrow \Omega _N$ be a bounded function which satisfies $\psi (\eta )=\psi (\eta ^{z,z+1})$ for any $z=x+1,\cdots , x+\varepsilon N-1$. For any $t\in [0,T]$ and $x\in \Lambda _N$ such that $x\in \{1,\cdots , N-\varepsilon N-2\}$ we have that

$$\begin{aligned} \begin{aligned}&{\limsup _{\varepsilon \rightarrow 0}}\displaystyle \limsup _{N\rightarrow +\infty } \mathbb {E}_{\mu _{N}}\left[ \, \left| \displaystyle \int _{0}^{t}\psi (\eta _{sN^2})(\eta _{sN^2}(x)- \overrightarrow{\eta }^{\varepsilon N}_{sN^2}(x))ds \right| \,\right] = 0. \end{aligned} \end{aligned}$$

(45)

Note that for $x\in \Lambda _N$ such that $x\in \{N-\varepsilon N-1, N-1\}$ the previous result is also true, but we replace in the previous expectation $\overrightarrow{\eta }^{\varepsilon N}_{sN^2}(x)$ by $\overleftarrow{\eta }^{\varepsilon N}_{sN^2}(x)$, where both averages were defined in (17).

Proof

We present the proof for the case when $x\in \{1,\cdots , N-\varepsilon N-2\}$ but we note that the other case is completely analogous. By applying the same arguments as in the proof of the previous lemma and by changing to the Bernoulli product measure $\nu _\alpha ^N$ with $\alpha \in (0,1)$, we can bound from above the previous expectation by

$$\begin{aligned} \begin{aligned} \tfrac{{C_{\alpha }}}{B} + t\sup _{f} \Big \{{ |}\langle \psi (\eta )(\eta (x)-\overrightarrow{\eta }^{\varepsilon N}(x)), f \rangle _{\nu ^N_\alpha } {|}+ \tfrac{N}{B} \langle {\mathcal L}_{N}\sqrt{f},\sqrt{f} \rangle _{\nu ^N_\alpha } \Big \}. \end{aligned} \end{aligned}$$

(46)

where B is a positive constant. The supremum above is over densities f with respect to $\nu _\alpha ^N$. The first term in the supremum above can be bounded by

$$\begin{aligned} \frac{1}{\varepsilon N}\sum _{y=x+1}^{x+\varepsilon N}{|}\langle \psi (\eta )(\eta (x)-\eta (y)), f \rangle _{\nu ^N_\alpha } {|.} \end{aligned}$$

By Lemma A.2 with the choice $A=\tfrac{B}{N}$ and from (42), the term on the right-hand side of (46), is bounded from above by ${B\varepsilon } +\tfrac{1}{N}.$ Taking $N \rightarrow \infty$, ${\varepsilon \rightarrow 0}$ and then $B\rightarrow +\infty$ we are done.$\square$

Now we state the replacement lemma that we need when the process is speeded up in the subdiffusive time scale.

Corollary A.5

Recall from (24) the definition of the mass of the system $m^N_t$ at the subdiffusive time scale $tN^{1+\theta }$. For any $\theta >1$ and $t\in [0,T]$ and $x\ne z\in \Lambda _N$ we have that

$$\begin{aligned} \begin{aligned}&\displaystyle \limsup _{N\rightarrow +\infty } \mathbb {E}_{\mu _{N}}\left[ \, \left| \displaystyle \int _{0}^{t}\eta _{sN^{1+\theta }}(z)(\eta _{sN^{1+\theta }}(x)-m_s^N )ds \right| \,\right] = 0. \end{aligned} \end{aligned}$$

(47)

The proof follows exactly the same strategy as above, the only difference being that when we use Lemma A.2 the function $\varphi (\eta )=\eta (z)$ is not invariant under the exchanges in the bulk. Nevertheless, by observing that the integrand function above can be written as

$$\begin{aligned} \begin{aligned} \eta (z)(\eta (x)-\langle \pi ^N,1\rangle )&=\frac{\eta (z)}{N-1}\sum _{y\ne x}(\eta (x)-\eta (y))\\&=\frac{\eta (z)}{N-1}(\eta (x)-\eta (z))+\frac{\eta (z)}{N-1}\sum _{y\ne x,z}{(\eta (x)-\eta (y))} \end{aligned} \end{aligned}$$

(48)

and thanks to the exclusion rule the first term in the last line of last display vanishes as $N\rightarrow +\infty$, and to finish the proof it is enough to estimate the second term in the last line of last display. To finish the proof, we distinguish two cases: either $x<z$ or $x>z$. We do the proof for the case $x<z$, but the other one is completely analogous. Observe that

$$\begin{aligned} \begin{aligned} \frac{\eta (z)}{N-1}\sum _{y\ne x,z}{(\eta (x)-\eta (y))}=\frac{\eta (z)}{N-1}\sum _{y=1}^{z-1}{(\eta (x)-\eta (y))}+\frac{\eta (z)}{N-1}\sum _{y=z+1}^{N-1}{(\eta (x)-\eta (y))}. \end{aligned} \end{aligned}$$

Now we explain how to estimate each one of the terms in the previous display. We present the proof for the first term but the second one is analogous. Therefore, we have to estimate

$$\begin{aligned} \begin{aligned}&\mathbb {E}_{\mu _{N}}\left[ \, \left| \displaystyle \int _{0}^{t}\frac{\eta _{sN^{1+\theta }}(z)}{N-1}\sum _{y=1}^{z-1}(\eta _{sN^{1+\theta }}(x)-\eta _{sN^{1+\theta }}(y))ds \right| \,\right] . \end{aligned} \end{aligned}$$

Now, we mimic the proof of the previous lemma. By following the first part of the proof, last expectation is bounded from above by

$$\begin{aligned} \begin{aligned} \tfrac{{C_{\alpha }}}{B} + t\sup _{f} \Big \{{ |}\langle \frac{\eta (z)}{N-1}\sum _{y=1}^{z-1}{(\eta (x)-\eta (y))}, f \rangle _{\nu ^N_\alpha } {|}+ \tfrac{N^\theta }{B} \langle {\mathcal L}_{N}\sqrt{f},\sqrt{f} \rangle _{\nu ^N_\alpha } \Big \}, \end{aligned} \end{aligned}$$

where B is a positive constant. The supremum above is over densities f with respect to $\nu _\alpha ^N$. Now, repeat the proof of Lemma A.2 and the previous lemma and to conclude, make the choice $A=BN^{-\theta }$. We leave the details to the reader.

Appendix B: Energy Estimate

Now we prove that the density $\rho (t,u)$ belongs to the space $L^{2}(0,T;{\mathcal {H}}^{1})$, see Definition 2.1. Define the linear functional $\ell _{\rho }$ defined in $C^{0,1}_{c}([0,T]\times (0,1))$ by

$$\begin{aligned} \ell _{\rho }(G) = \int ^{T}_{0}\int ^{1}_{0}\partial _uG_{s}(u)\rho (s,u) \, du ds = \int ^{T}_{0}\int ^{1}_{0}\partial _uG_{s}(u)\, \pi (s,du) ds. \end{aligned}$$

Proposition B.1

There exist positive constants C and c such that

$$\begin{aligned} \mathbb {E}\Big [ \sup _{G\in C^{0,1}_{c}([0,T]\times (0,1))}\Big \{ \ell _{\rho }(G) - c \Vert G \Vert _{2}^{2}\Big \}\Big ]\leqslant C < \infty . \end{aligned}$$

Above $\Vert G \Vert _{2}$ denotes the norm of a function $G \in L^{2}([0,T]\times (0,1)).$

Before proving this result, we state and prove an energy estimate for the macroscopic density.

Corollary B.2

Any limit point $\mathbb {Q}$ of the sequence $(\mathbb {Q}_N)_{N\geqslant 1}$ satisfies

$$\begin{aligned} \mathbb {Q}\left( \pi _\cdot \in \mathcal {D}([0,T],\mathcal {M}), \;\;\pi _t:=\rho _t(u) du, \;\;\rho \in L^2(0;T,{\mathcal {H}}^1)\right) =1. \end{aligned}$$

We denote $R_T$ the event above.

Proof of Corollary B.2

We first note that because of the exclusion between particles, every limit point $\mathbb {Q}$ is concentrated on trajectories of measure that are absolutely continuous with respect to the Lebesgue measure (see e.g. [17], p.57, last paragraph for more details).

From Proposition B.1, $\ell _{\rho }$ is ${\mathbb Q}$-almost surely continuous and therefore we can extend this linear functional to $L^{2}([0,T]\times (0,1))$. As a consequence of the Riesz’s Representation Theorem there exists $H \in L^{2}([0,T]\times (0,1))$ such that

$$\begin{aligned} \ell _{\rho }(G) = -\int ^{T}_{0}\int ^{1}_{0} G_{s}(u)H_{s}(u)du ds \end{aligned}$$

for all $G \in C^{0,1}_{c}([0,T]\times (0,1))$. From this we conclude that $\rho \in L^{2}(0,T;{\mathcal {H}}^{1})$.$\square$

Proof of Proposition B.1

By density and by the Monotone Convergence Theorem it is enough to prove that for a countable dense subset $\lbrace G_{m}\rbrace _{m \in \mathbb {N}}$ on $C_{c}^{0,2}([0,T]\times (0,1))$ it holds that

$$\begin{aligned} \mathbb {E}\left[ \max _{k \leqslant m}\lbrace \ell _{\rho }(G^{k}) - c \Vert G^{k} \Vert _{2}^{2}\rbrace \right] \leqslant C_{0}, \end{aligned}$$

for any m and for $C_{0}$ independent of m. Note that the function that associates to a trajectory $\pi _\cdot \in {\mathcal D}([0,T], {\mathcal M}^+)$ the number $\max _{k\leqslant m}\left\{ \ell _{\rho }(G^{k})- c\Vert G^{k} \Vert _{2}^{2} \right\} ,$ is continuous and bounded w.r.t. the Skorohod topology of ${\mathcal D}([0,T], {\mathcal M}^+)$ and for that reason, the expectation in the previous display is equal to the next limit

$$\begin{aligned} \lim _{N\rightarrow \infty } {\mathbb {E}}_{\mu _{N}}\left[ \max _{k\leqslant m}\left\{ \int _{0}^{T}\dfrac{1}{N-1} \sum _{x=1}^{N-1} \partial _{u}G_{s}^{k}(\tfrac{x}{N})\eta _{{sN^2}}(x)ds -c\Vert G^{k} \Vert _{2}^{2} \right\} \right] . \end{aligned}$$

By entropy and Jensen’s inequalities plus the fact that $e^{\max _{k\leqslant m} a_{k}}\leqslant \sum _{k=1}^{m}e^{a_{k}}$ the previous display is bounded from above by

$$\begin{aligned}\nonumber&C_{0} + \dfrac{1}{N}\log \mathbb {E}_{\nu ^N_\alpha }\left[ \sum _{k =1}^{m}{\exp \Big \{{\int _{0}^{T}\sum _{x \in \Lambda _{N}}\partial _{u}G_{s}^{k}(\tfrac{x}{N})\eta _{{sN^2}}(x) ds - cN\Vert G^{k} \Vert _{2}^{2} }\Big \}} \right] ,\\ \end{aligned}$$

By linearity of the expectation, to treat the second term in the previous display it is enough to bound the term

$$\begin{aligned} \limsup _{N \rightarrow \infty } \; \dfrac{1}{N}\log \mathbb {E}_{\nu ^N_\alpha }\left[ {\exp \Big \{\int _{0}^{T}\sum _{x \in \Lambda _{N}}\partial _{u}G_{s}(\tfrac{x}{N})\eta _{{sN^2}}(x) ds - cN\Vert G \Vert _{2}^{2} \Big \}}\right] , \end{aligned}$$

for a fixed function $G\in C_{c}^{0,2}([0,T]\times (0,1))$, by a constant independent of G. By Feynman-Kac’s formula, the expression inside the limsup is bounded from above by

$$\begin{aligned} \int _{0}^{T}\sup _{f}\Big \{\dfrac{1}{N}\int _{\Omega _{N}}\sum _{x \in \Lambda _{N}}\partial _{u}G_{s}(\tfrac{x}{N})\eta (x) f(\eta )d{\nu ^N_{\alpha } } - c\Vert G \Vert _{2}^{2} + N \langle {\mathcal L}_{N}\sqrt{f},\sqrt{f}\rangle _{{\nu ^N_{\alpha }}}\Big \}\, ds{,} \end{aligned}$$

(49)

where the supremum is carried over all the densities f with respect to $\nu ^N_{\alpha }$. Note that by a Taylor expansion on G, it is easy to see that we can replace its space derivative by the discrete gradient $\nabla ^+_{N} G_{s}(\tfrac{x-1}{N})$ by paying an error of order $O(\frac{1}{N})$. Then, from a summation by parts, we obtain

$$\begin{aligned} \int _{\Omega _{N}} \sum _{x=1}^{N-2} G_{s}(\tfrac{x}{N})(\eta (x)-\eta (x+1))f(\eta )d\nu ^N_{\alpha } \end{aligned}$$

By writing the previous term as one half of it plus one half of it and in one of the halves we swap the occupation variables $\eta (x)$ and $\eta (x+1)$, for which the measure $\nu _\alpha$ is invariant, the last display becomes equal to

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\int _{\Omega _{N}} \sum _{x=1}^{N-2} G_{s}(\tfrac{x}{N})(\eta (x)-\eta (x+1))(f(\eta )-f(\eta ^{x,x+1}))d{\nu ^N_{\alpha }}. \end{aligned} \end{aligned}$$

(50)

Repeating similar arguments to those used in the proof of Lemma A.2, the last term is bounded from above by

$$\begin{aligned}&\dfrac{1}{4N}\int _{\Omega _{N}} \sum _{x=1}^{N-2} (G_{s}(\tfrac{x}{N}))^{2}(\sqrt{f(\eta )}+ \sqrt{ f(\eta ^{x,x+1})})^{2}d\nu ^N_{\alpha }\\&+\dfrac{1}{4N}\int _{\Omega _{N}} \sum _{x=1}^{N-2} (\sqrt{f(\eta )}- \sqrt{ f(\eta ^{x,x+1})})^{2}d\nu ^N_{\alpha } \nonumber \\&\quad \leqslant \dfrac{C }{N}\sum _{x\in \Lambda _{N}} (G_{s}(\tfrac{x}{N}))^{2}+ \dfrac{1}{4N} D_{0,N}(\sqrt{f},\nu ^N_{\alpha }) \end{aligned}$$

for some $C>0$. From (42) we get that (49) is bounded from above by

$$\begin{aligned} C \int _0^T \Big [ 1 + \dfrac{1}{N} \sum _{x\in \Lambda _{N}} (G_{s}(\tfrac{x}{N}))^{2}\Big ] \, ds \; -\; c \Vert G\Vert _2^2 \end{aligned}$$

plus an error of order $O(\tfrac{1}{N})$. Above C is a positive constant independent of G. Since $\dfrac{1}{N} \sum _{x\in \Lambda _{N}} (G_{s}(\tfrac{x}{N}))^{2}$ converges, as $N\rightarrow +\infty$, to $\Vert G\Vert _2^2$, then it is enough to choose $c>C$ to conclude that

$$\begin{aligned} \limsup _{N\rightarrow \infty } \; \Big \{ C \int _0^T \Big [ 1 + \dfrac{1}{N} \sum _{x\in \Lambda _{N}} (G_{s}(\tfrac{x}{N}))^{2}\Big ] \, ds \; -\; c \Vert G\Vert _2^2 \Big \} \; \lesssim \; 1 \end{aligned}$$

and we are done.$\square$

Appendix C: Uniqueness of Weak Solutions of (8)

We start this section by recalling from Sect. 7.2 of [4] the next two lemmas, which will be used in our proof. The first one concerns uniqueness of the strong solutions of the heat equation with linear Robin boundary conditions.

Lemma C.1

For any $t\in (0,T]$, the following problem with Robin boundary conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _s\varphi (s,u)+a\partial ^2_u\varphi (s,u)=\lambda \varphi (s,u),&{}(s,u)\in [0,t{)}\times (0,1),\\ \partial _u \varphi (s,0)=b(s,0)\varphi (s,0), &{} s\in [0,t),\\ \partial _u \varphi (s,1)=-b(s,1)\varphi (s,1), &{} s\in [0,t),\\ \varphi (t,u)=h(u), &{} u\in (0,1), \end{array}\right. } \end{aligned}$$

(51)

with $h\equiv h(u)\in C_0^2([0,1])$ , $\lambda \geqslant 0$ , $0<a\equiv a(u,t)\in C^{2,2}([0,T]\times [0,1])$, and for $u\in \{0,1\}$, $0<b\equiv b(u,t)\in C^2[0,T]$, has a unique solution $\varphi \in C^{1,2}([0,t]\times [0,1])$. Moreover, if $h\in [0,1]$ then we have $\forall (s,u)\in [0,t]\times [0,1]$:

$$\begin{aligned} 0\leqslant \varphi (s,u)\leqslant e^{-\lambda (t-s)}. \end{aligned}$$

The second lemma is a technical regularization result on the coefficients $b(s,\cdot ).$

Lemma C.2

Let $0\leqslant b$ be a bounded measurable function in [0, T] , $A=\{t\in [0,T]:b(t)>0\}$ and $p\in [1,\infty )$. Then there is a sequence $(b_k)_{k\geqslant 0}$ of positive functions in $C^{\infty }[0,T]$ such that $b_k\xrightarrow {k\rightarrow \infty }b$ in $L^p([0,T])$ and

$$\begin{aligned} \left\| \frac{b}{b_k}-1\right\| _{L^{p}(A)}\xrightarrow {k\rightarrow \infty }0. \end{aligned}$$

For the proof of Lemma 2.4, that is of uniqueness of weak solutions of (8), we will follow Filo’s method [12], but mostly as presented in Sect. 7.2 of [4]. The main idea is to choose a particular test function for the weak formulation satisfied by $w:=\rho ^{(1)}-\rho ^{(2)}$, where $\rho ^{(1)}$ and $\rho ^{(2)}$ are two weak solutions with the same initial data. Although we do not have as much work to treat the bulk terms as in [4], our main issue is the non linearity of the boundary conditions.

Recalling the weak formulation in (9) and Lemma 5.4, since

$$\begin{aligned} D_{\lambda ,\sigma }\rho _s^{(1)}(v)-D_{\lambda ,\sigma }\rho _s^{(2)}(v)=-w_s(v) V_{\lambda ,\sigma }(\rho _s^{(1)},\rho _s^{(2)})(v,v):=-w_s(v)V_{\lambda ,\sigma }(v,s) \end{aligned}$$

for $v=0,1$ and $(\lambda ,\sigma )=(\alpha ,\gamma ),(\beta ,\delta )$, we have:

$$\begin{aligned} \langle w_{t}, G_{t} \rangle = \int _0^t\langle w_{s},\Big ( \partial ^2_u + \partial _s\Big ) G_{s} \rangle ds + \int _0^t w_{s}(0)\Big ( \partial _u G_{s}(0) - G_s(0) V_{\alpha ,\gamma }(0,s) \Big ) ds\nonumber \\ -\int _0^t w_{s}(1) \Big ( \partial _u G_{s}(1) + G_s(1) V_{\beta ,\delta }(1,s) \Big ) ds. \end{aligned}$$

(52)

Now we choose our test functions. Since $V_{\cdot ,\cdot }$ does not have enough regularity, we have to overcome this problem by using Lemma C.2. We focus on the left boundary, since for the right boundary the computations are analogous. Let $A_0=\{t\in [0,T]: V_{\alpha ,\gamma }(0,t)>0 \}$ (similarly, we define $A_1$ with respect to the right boundary). From Lemma 5.4 we have $V_{\alpha ,\gamma }(0,s)>0$ and we may therefore exchange [0, t] by $A_0$ (resp. $A_1$) and apply Lemma C.2. As a consequence of Lemma C.2, for k large enough, there exists $b_{k}(s,0)$ close to $V_{\alpha ,\gamma }(0,s)$ in $L^p([0,T])$ for $p\in [1,+\infty )$:

$$\begin{aligned} \left\| \frac{ V_{\alpha ,\gamma }(0,\cdot )}{b_k(\cdot ,0)}-1\right\| _{L^p(A_0)}\leqslant \epsilon \end{aligned}$$

for $\epsilon >0$ and $A_0=\{ s\in [0,t]: V_{\alpha ,\gamma }(0,s)>0 \}$. Now we choose the space of test functions as a sequence $\varphi _k$, where for each k, the function $\varphi _k$ solves (51) with $b_k(\cdot ,0)$ given above and with $\lambda =0$. From the boundary conditions of (51), second term in (52) writes as

$$\begin{aligned}&\int _0^t \varphi _k(0,s)w_s(0) \left( b_k(0,s)- V_{\alpha ,\gamma }(0,s) \right) ds. \end{aligned}$$

(53)

Exchanging [0, t] by $A_0$, the last display can be bounded from above by

$$\begin{aligned}&\Big |\int _{A_0} \varphi _k(0,s)w_s(0)b_k(0,s) \left( 1-\frac{ V_{\alpha ,\gamma }(0,s)}{b_k(0,s)} \right) ds\Big |\\&\quad \leqslant 2\left\| b_k(0,s)\right\| _{L^1(A_0)}\left\| \frac{ V_{\alpha ,\gamma }(0,s)}{b_k(0,s)}-1\right\| _{L^1(A_0)}\lesssim \epsilon , \end{aligned}$$

where we used that $\varphi _k(\cdot )$ and $w(\cdot )$ are bounded functions. For the right boundary the argument is completely analogous.

Now we treat the bulk term. From our choice of test function we have

$$\begin{aligned} \int _0^t\langle w_s,(\partial ^2_u+\partial _s)\varphi _k \rangle ds =\int _0^t\langle w_s,(1-a)\partial ^2_u \varphi _k(\cdot ,s) \rangle ds. \end{aligned}$$

Letting $a=1$, we thus have that $\langle w_t,\varphi _t \rangle \lesssim \epsilon$. Since $\varphi _t=h$, it is enough to take $h\equiv h_k\in C_0^2([0,1])$ such that $h_k(\cdot )\xrightarrow {k\rightarrow \infty }1_{\{u\in [0,1]:w_t(u)>0 \}}(t,\cdot )$ in $L^2([0,1])$. The conclusion follows straightforwardly.

Remark C.3

We remark that for the model in [5] the lower bound takes the form of the term $x=K$ in the sum above, with $\gamma =0$. For $K=1$, that is, the case studied in [1], we have $V=\rho ^{(1)}+\rho ^{(2)}$, and thus $V=0\implies w=0$.